What is the average rate of change of the function $g(x)=\sin(x)$ over the interval $1\le x \le1+h$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin\left(1+h\right)-\sin(1)}{1+h}$ (Choice B) B $\dfrac{\sin\left(1+h\right)-\sin\left(1\right)}{h}$ (Choice C) C $\dfrac{\sin(h)-\sin\left(1\right)}{h}$ (Choice D) D $\dfrac{\sin\left(1+h\right)-\sin(1)}{1}$
Solution: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We are interested in the average rate of change of $g(x)=\sin(x)$ over the interval $1\le x \le1+h$ : $\begin{aligned} &\phantom{=}\dfrac{g(1+h)-g(1)}{(1+h)-(1)} \\\\ &=\dfrac{\sin(1+h)-\sin(1)}{1+h-1} \\\\ &=\dfrac{\sin(1+h)-\sin(1)}{h} \end{aligned}$ The average rate of change of the function is $\dfrac{\sin(1+h)-\sin(1)}{h}$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.